Add solution for Project Euler 65, Fixes: TheAlgorithms#2695

Author: peteryao7

Diff for: DIRECTORY.md

@@ -652,6 +652,8 @@
     * [Sol1](https://github.com/TheAlgorithms/Python/blob/master/project_euler/problem_56/sol1.py)
   * Problem 63
     * [Sol1](https://github.com/TheAlgorithms/Python/blob/master/project_euler/problem_63/sol1.py)
+  * Problem 65
+    * [Sol1](https://github.com/TheAlgorithms/Python/blob/master/project_euler/problem_65/sol1.py)
   * Problem 67
     * [Sol1](https://github.com/TheAlgorithms/Python/blob/master/project_euler/problem_67/sol1.py)
   * Problem 76

Diff for: project_euler/problem_65/__init__.py

@@ -0,0 +1,90 @@
+"""
+The square root of 2 can be written as an infinite continued fraction.
+
+sqrt(2) = 1 + 1 / (2 + 1 / (2 + 1 / (2 + 1 / (2 + ...))))
+
+The infinite continued fraction can be written, sqrt(2) = [1;(2)], (2)
+indicates that 2 repeats ad infinitum. In a similar way, sqrt(23) =
+[4;(1,3,1,8)].
+
+It turns out that the sequence of partial values of continued
+fractions for square roots provide the best rational approximations.
+Let us consider the convergents for sqrt(2).
+
+1 + 1 / 2 = 3/2
+1 + 1 / (2 + 1 / 2) = 7/5
+1 + 1 / (2 + 1 / (2 + 1 / 2)) = 17/12
+1 + 1 / (2 + 1 / (2 + 1 / (2 + 1 / 2))) = 41/29
+
+Hence the sequence of the first ten convergents for sqrt(2) are:
+1, 3/2, 7/5, 17/12, 41/29, 99/70, 239/169, 577/408, 1393/985, 3363/2378, ...
+
+What is most surprising is that the important mathematical constant,
+e = [2;1,2,1,1,4,1,1,6,1,...,1,2k,1,...].
+
+The first ten terms in the sequence of convergents for e are:
+2, 3, 8/3, 11/4, 19/7, 87/32, 106/39, 193/71, 1264/465, 1457/536, ...
+
+The sum of digits in the numerator of the 10th convergent is
+1 + 4 + 5 + 7 = 17.
+
+Find the sum of the digits in the numerator of the 100th convergent
+of the continued fraction for e.
+"""
+
+
+def solution(max: int) -> int:
+    """
+    The solution mostly comes down to finding an equation that will generate
+    the numerator of the continued fraction. For the i-th numerator, the
+    pattern is:
+
+    n_i = m_i * n_(i-1) + n_(i-2)
+
+    for m_i = the i-th index of the continued fraction representation of e,
+    n_0 = 1, and n_1 = 2.
+
+    For example:
+    n_9 = 6 * 193 + 106 = 1264
+    1 + 2 + 6 + 4 = 13
+
+    n_10 = 1 * 193 + 1264 = 1457
+    1 + 4 + 5 + 7 = 17
+
+    >>> solution(9)
+    13
+    >>> solution(10)
+    17
+    """
+    n0 = 1
+    n1 = 2
+
+    for i in range(2, max + 1):
+        temp = n0
+        m = 2 * i // 3 if i % 3 == 0 else 1
+        n0 = n1
+        n1 = m * n0 + temp
+
+    return sum_digits(n1)
+
+
+def sum_digits(num: int) -> int:
+    """
+    Adds all the single digits of an int together.
+
+    >>> sum_digits(1)
+    1
+    >>> sum_digits(12345)
+    15
+    >>> sum_digits(999001)
+    28
+    """
+    digit_sum = 0
+    while num > 0:
+        digit_sum += num % 10
+        num //= 10
+    return digit_sum
+
+
+if __name__ == "__main__":
+    print(solution(100))